What is the general solution for 2cos(x)=cos^2(x)-1 ?

The general solution for 2cos(x)=cos^2(x)-1 is x=1.99787…+2pin,x=-1.99787…+2pin

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2cos(x)=cos^2(x)-1

2 cos (x) = cos^{2} (x) - 1

x = 1.99787 \dots + 2 πn, x = - 1.99787 \dots + 2 πn

Solution steps

2 cos (x) = cos^{2} (x) - 1

Solve by substitution

cos (x) = 1 + 2, cos (x) = 1 - 2

cos (x) = 1 + 2 :

No Solution

cos (x) = 1 - 2 : x = arccos (1 - 2) + 2 πn, x = - arccos (1 - 2) + 2 πn

Combine all the solutions

x = arccos (1 - 2) + 2 πn, x = - arccos (1 - 2) + 2 πn

Show solutions in decimal form

x = 1.99787 \dots + 2 πn, x = - 1.99787 \dots + 2 πn

2 sin (v) csc (v) - csc (v) = 4 sin (v) - 2

2 cos (2 x) = 4 cos (x)

\frac{sin ( 4 2 ^{\circ} )}{22} = \frac{sin ( B )}{12}

sin (θ) = \frac{16}{14}

tan (θ) = \frac{83}{47}

What is the general solution for 2cos(x)=cos^2(x)-1 ?
The general solution for 2cos(x)=cos^2(x)-1 is x=1.99787…+2pin,x=-1.99787…+2pin